Questions: 1 2 3 4 5 6 7
Physics 1120: Moment of Inertia Solutions
1. The moment of inertia of an oxygen molecule about an axis through the centre of mass and perpendicular
to the line joining the atoms is 1.95 × 10­46 kg­m2. The mass of an oxygen atom is 2.66 × 10­26 kg. What
is the distance between the atoms? Treat the atoms as particles.
The moment of inertia for point particles is given by Rewriting this for L yields .
.
2. An empty beer can has a mass of 50 g, a length of 12 cm, and a radius of 3.3 cm. Assume that the shell of
the can is a perfect cylinder of uniform density and thickness.
(a) What is the mass of the lid/bottom?
(b) What is the mass of the shell?
(c) Find the moment of inertia of the can about the cylinder's axis of symmetry.
(a) The mass has three components, M = Mlid + Mshell + Mbottom = 50 g, where the lid and the bottom
are identical. Mass is proportional to the surface area for uniform objects. The area of the lid and bottom
is that of a circle, Acircle = πr2 . The surface area of a cylindrical shell is Ashell = 2πrL. So the total area of
can is Atotal = 2Acircle + Ashell = 2πrL + 2πr2 = 2πr(r + L). So the mass of the lid or bottom is given by . Thus Mlid = 5.392 g.
(b) Similarly, the mass of the shell is given by . Thus Mshell = 39.216 g.
(c) The total moment of inertia of the beer can is given by the sum of the individual pieces, Itotal = Ilid +
Ishell + Ibottom. Since each piece is revolving about its centre of mass, we do not need the parallel axis
theorem. Looking up the moments of inertia of a flat solid disk and a thin cylindrical shell, we have Itotal =
2×½Mlidr2 + Mshellr2 = 4.86 × 10­5 kg­m2.
3. A dumbbell consists of two uniform spheres of mass M and radius R joined by a thin rod of mass m,
length L, and radius r (see diagram).
(a) What is the moment of inertia about the centre of mass and perpendicular to the rod (Axis A)?
(b) About an axis through one sphere and perpendicular to the rod (Axis B)?
(c) About an axis along the rod (Axis C)? (a) The moment of inertia of a composite body is equal to the sum of the moments of its individual pieces, .For axis A, the rod is rotating about its centre of mass. Each sphere is a
distance R+L/2 from the axis of rotation, so we must use the parallel axis theorem. Recall that the moment
of inertia of a rod about its centre is and that the moment of inertia of a solid sphere about its
centre is . Thus we have
(b) For axis B, the rod's centre is R+L/2 away from the axis of rotation. One sphere's centre is L+2R
from the axis of rotation. The last sphere is rotating about axis B. Thus r
(c) The bar and both spheres are rotating about their own centres when rotating about axis c. However,
note that the bar is a cylinder or radius r in this configuration.
4. A hole of radius r has been drilled in a flat, circular plate of radius R. The centre of the hole is at a distance
d from the centre of the circle. The mass of the complete body was M. Find the moment of inertia for
rotation about an axis through the centre of the plate.
We must treat the hole as an object of negative mass. The inertia of the object is then just
. The plate and the hole are just disks and the inertia of a disk is . The hole is
not rotating about its own centre of mass, so we must use the parallel axis theorem,
.
We are not yet finished; we do not know the mass of the hole. However since the plate is uniform, mass is
proportional to area, so
.
Substituting this into our equation for the moment of inertia yields
.
5. A thick spherical shell has an inner radius R1, an outer radius R2, and a mass M. The material that the
spherical shell is made of is uniform. Find the moment of inertia of the thick shell about an axis through the
centre of the sphere. The volume of a sphere is 4πr3/3.
We must treat the hole at the centre as a sphere of negative mass. Since the moment of inertia of a sphere
about its centre is , the moment of inertia of this object is
.
We are not yet finished since we do no know the mass of either the sphere or of the hole. We must find a
way to relate these masses to the mass of the spherical shell, M. Fortunately the object is uniform so the
mass of each is related to its volume. The volume of the spherical shell is simply the difference of its outer
and inner volumes,
.
Thus the mass of the sphere is given by
.
Similarly, the mass of the hole is given by
.
Thus the moment of inertia of a thick spherical shell is
.
6. The object shown in the diagram below consists of a 100­kg, 25.0 cm radius cylinder connected by four
5.00­kg, 0.75­m long thin rods to a thin­shelled outer cylinder of mass 20.0 kg. A small chunk of metal of
mass 1.00 kg has been welded to the outer cylinder. What is the moment of inertia of the object about the
centre of the inner cylinder? Treat the metal chunk as a point mass.
The moment of inertia of a composite body is equal to the sum of the moments of its individual pieces, . The rods are not rotating about their centre of mass, so we must use
the parallel axis theorem. The centre of each rod is Rcyl+L/2 from the axis of rotation at the centre of the
object. The moments of inertia for a cylindrical shell, a disk, and a rod are MR2, respectively. The moment of inertia of a point mass is , and . Thus the total moment of inertia is:
.
7. The object in the diagram below consists of five thin cylinders arranged in a circle. A thin disk has been
welded to the tops and the bottoms of the cylinders. The cylinders each have a mass of 250 g, a length of
15.0 cm, and a radius of 1.00 cm. The disks are each 125 g and have a radius of 5.00 cm. Find the
moment of inertia of the whole object about an axis through the centres of the disks. The moment of inertia of a composite body is equal to the sum of the moments of its individual pieces, . The cylinders are not rotating about their centre of mass, so we must use the
parallel axis theorem. The centre of each rod is Rdisk­ Rcyl from the axis of rotation at the centre of the
object. The moments of inertia for a a disk or a cylindrical rod are is . The moment of inertia of a
point mass is . Thus the total moment of inertia is:
.
Questions? mike.coombes@kwantlen.ca